Definition
A first order differential equation is called separable if the derivative of $y$ can be expressed as the product of a function dependent on $t$ and function dependent on $y$:
\[\frac{dy}{dt}=f(t,y)=g(t)h(y)\]where both $g(t)$ and $h(y)$ are continuous functions.
Autonomous Equations
Note that in the special case that $g(t)=1$, the separable equation is independent of $t$ and is thus a first order autonomous equation.
Solutions by Integration
The reason differential equations of the above form are interesting is because they are equivalent to the following (hopefully simple) integral equation:
\[\int\frac{1}{h(y)} \,dy=\int g(t) \,dt\]which, when computed, will yield a family of implicit solutions to the differential equation (which can possibly be solved for a family of explicit solutions). We call this technique for solving differential equations the separation of variables.
Proof
$$\begin{align*} \frac{dy}{dt}&=g(t)h(y)\\ \frac{1}{h(y)}\frac{dy}{dt}&=g(t)&\text{(algebra)}\\ \int\frac{1}{h(y)}\frac{dy}{dt}\,dt&=\int g(t)\,dt&\text{(integrate with $dt$)}\\ \int\frac{1}{h(y)}\,dy&=\int g(t)\,dt&\text{(algebra)} \end{align*}$$ In the last step we multiply $\frac{dy}{dt}$ by $dt$ to get $dy$. This can be more directly justified using non-standard analysis where infinitesimal quantities are meaningful.Equilibrium Solutions
Note that the second step of our proof (where we divided both sides by $h(y)$) is not valid for constant $y_0$ such that $h(y_0)=0$ as division by zero is undefined. As such, we must take special care in checking the roots of $h(y)$. In particular, we note that these roots are actually solutions to the separable equation since the derivative of a constant is $0$:
\[\frac{d}{dt}y_0=0\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{dy}{dt}=g(t)h(y_0)=0\]For any constant $y_0$ such that $h(y_0)=0$.
These constant solutions for $y$, called equilibrium solutions, aren’t present in the solution to the integral equation above. And so the entire solution set of a separable equation consists of the roots of $h(y)$ as well as the solutions of the integral equation.